Question

In an acidic solution $I^{-}$ changes to $I_2$. How many grams of $I_2$ is produced if, in the same process, $1.5\times {10}^{22}$ electrons are used up to reduce $H_{3}As{O}_4$ to $H_{3}As{O}_3$? ($I=127$)

• A. $1.6g$
• B. $6.4g$
• C. $4.8g$
• D. $3.2g$

Answer 1

The correct option is D $3.2g$

$\begin{array}{c}\text{Given:}-2I^{-}⟶I_2+2e^{-}-\left(1\right)\\ \text{For same process.}\\ As{O}_4^{3-}⟶As{O}_3^{3-}\\ \text{using}1.5\times {10}^{22}\text{electrons}\\ \text{These, electrons should be provided by}\\ \text{reaction (1)}\\ \text{moles of electrons}=\frac{1.5\times {10}^{22}}{6.022\times {10}^{23}}=0.02491\end{array}$

Now, 2 mole I- produce 1 mole $I_2$ and 2 moles

$\therefore 0.02491$ the produce 0.02491 mole $e^{-}$

and moles of $I_2=\frac{0.02491}{2}$

$\therefore$ grams of $I_2$ produced $=\frac{12.7\times 2\times 0.02491}{2}$

$=3.164g$

$I_2\approx 3.2g$

So, option $D$ is correct.